Subnetting Practice Question Types

Question Type 1

What is the Network ID, Broadcast Address, First Usable IP, or Last Usable IP on the subnet that the node belongs to?

No matter which of the four criteria the question asks, you should always follow these identical steps.

Step 1. Convert the shorthand subnet mask to decimal.

  • /26 = 255.255.255. + Two additional subnet bits.
  • Recall the left-most two bits are the 128 place and 64 place. 128+64 = 192
  • Thus, our decimal subnet mask is

Step 2. Determine the block size.

  • The block size in that octet is 256-decimal
  • 256 – 192 = 64
  • The block size is 64.

Step 3. What is my Network ID?

  • Since we are working in the fourth octet and the block size is 64, the first network is

Step 4. What is the next Network ID?

  • Again, we look at our octet block size of 64 to determine the next network is
  • Network ID (First IP in the subnet):
  • Broadcast address (last IP in the subnet):
  • First Usable IP (the address after the network ID):
  • Last Usable IP (the address before the broadcast address):

Notes: The block size determines the networks. Since the block size is 64, our networks are,,, and To determine your network, you just have to find the range that includes your IP address. In this example, it is the first network. If your IP was, you would belong to the second network –

Question Type 2

You have been asked to create a subnet mask for the network. Your organization requires 900 subnets, with at least 50 hosts per subnet. What subnet mask should you use?

Step 1: Determine how many network bits (1’s), you have to add to the classful boundary to cover the number of required subnets.

  • The IP address given is a class B private address, making the first 16 bits network bits for our classful subnetting.
  • Find the exponent of 2 that is equal to or greater than the number of subnets we require (900). 10 additional subnet bits will give us 1,024 subnets. Add 10 1’s to the /16 class B subnet mask, making for /26 subnet mask. Make note of the corresponding subnet mask. In this case, The third octet is eight 1’s, and the fourth is two 1’s.

Step 2: Confirm the number of remaining host bits will cover our required hosts. In this case, there are 6 remaining 0’s in our subnet mask, 2^6-2=62, which is more than enough for our host requirement of 50.

Our subnet mask is Giving us 1024 subnets and 62 hosts per subnet.

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